I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan

Jul 13th 2009, 03:18 AM

CaptainBlack

Quote:

Originally Posted by auonline

I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan

Yes the required area is:

what method you use to evaluate it is up to you and/or your teacher.

CB

Jul 13th 2009, 03:25 AM

auonline

Ok thanks :D

Jul 13th 2009, 05:58 AM

VonNemo19

Quote:

Originally Posted by auonline

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Another way:

Jul 13th 2009, 06:35 AM

HallsofIvy

Quote:

Originally Posted by VonNemo19

Another way:

Ah, the easy way! (Rofl)

Jul 13th 2009, 06:50 AM

HallsofIvy

Quote:

Originally Posted by auonline

I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan

If you let u= x-2, differentiating both sides give du= dx (was that what you meant by "finds the derivate"); when x= -1 , u= -1-2= -3, and when x= 1, u= 1-2= -1. Now replace each of those things in your integral by its value in terms of u, becomes . That is NOT some "random hard to understand process". That is a very logical, standard method known as "integration by substitution".

Jul 13th 2009, 02:16 PM

VonNemo19

Quote:

Originally Posted by HallsofIvy

Ah, the easy way! (Rofl)

I think so. I realize that this is not as quick as u-substitution, but it is good to see things from all sides when trying to tackle a problem.